# Need help with Physics problem (momentum/collisions)

1. Apr 7, 2014

### Engineering100

1. The problem statement, all variables and given/known data

A 0.800 kg target slides along the ice at 3.0 m/s [W], when it is hit by a 20.0 g arrow moving at 260 m/s [N], as part of a show. Find the final velocity of the target after the inelastic collision.

2. Relevant equations

pi= pf
m1 v1i+ m2 v2i = (m1+ m2)vf
pythagorean theorem

3. The attempt at a solution

v1f= ?

pix= pfx
m1 v1ix+ m2 v_2ix = (m1+ m2)vfx
(0.80)(-3.0)+0=(0.80+0.020) vfx
-2.4=(0.820)vfx
v_fx= -2.93 m/s

piy= pfy
m1 v1iy+ m2 v2iy = (m1+ m2)vfy
0+(0.02)(260)=(0.800+0.02) vfy
vfy=5.2/0.82=6.34 m/s

vf=√(-2.93)^2+(6.34)^2
vf=6.98 m/s

tanθ= -2.93/6.34
θ=25°

Since one of these values are negative, the angle is negative. Relatively speaking what would the angle be. also was this question in general answered correctly?

2. Apr 7, 2014

### electricspit

Yes you used the conservation of linear momentum properly, and successfully applied it to each coordinate individually.

The last step however is wrong, as the ratio should be:

$\tan{\theta}=\frac{6.34}{-2.93}$

Which would give:

$\theta=-65.2^\circ$

The negative angle interpretation depends on how you defined your angles. In this case, the angles are defined in the sense such that $\theta=0$ along the "negative x-axis". Therefore a negative angle just represents clockwise rotation as opposed to counter-clockwise rotation. So it is travelling $65^\circ$ in the North-West direction. Make sense?

Normally, for example, $\theta=45^\circ$ would be above the x-axis in the 1st quadrant, whereas $\theta=-45^\circ$ would be below the x-axis in the 4th quadrant. In this case $\theta=45^\circ$ would be in the 3rd quadrant below the x-axis and $\theta=-45^\circ$ would be above the x-axis in the second quadrant.

This is due to the fact that the target is moving initially at an angle $-180^\circ$ which is now taken as $0^\circ$. Does this help?

Last edited: Apr 7, 2014
3. Apr 7, 2014

### Engineering100

yes thanks so much, great help

4. Apr 7, 2014

### Engineering100

so would that be [W 65° N] or [N 25° W]?

5. Apr 7, 2014

### Staff: Mentor

Yes. You did it right, but why did you assume that the target was moving in the negative x direction to start with? There's nothing wrong with this, but I would have had it going in the + x direction (for some reason). Oh well. Potatoes, Potahtoes.

Chet

6. Apr 7, 2014

### electricspit

If I understand what you're saying correctly then it would be $\text{[W } 65^\circ \text{ N]}$ (65 degrees North of West).

Also he did it that way because it specified that it was travelling West, and in introductory physics no one expects people to think like that probably.

7. Apr 7, 2014

### Staff: Mentor

Oh thanks. I didn't notice that [W].

Chet

8. Apr 7, 2014

### Engineering100

yes, is [W 65∘ N] the correct angle?